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C 2 g, then row reduction of OE c 1 c 2 b 1 b 2 to OE I P produces a matrix P that satisfies OE x B D P OEF(x)dx is called the definite integral of f(x) over the interval a,b and stands for the area underneath the curve y = f(x) over the interval a,b (with the understanding that areas above the xaxis are considered positive and the areas beneath the axis are considered negative)G(x)=a, g(y)=d, g(z)=c, g(w)=b;

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Sep 30, 17 · Answer is (ab)/2 int_a^b f(x) dx Let "I" be the integral, I = int_a^b xf(x) dx (1) Using the property of definite integrals, I = int_a^b (abx) f(abx) dx It is given that, f (abx) = f(x) implies I = int_a^b (abx) f(x) dx implies I = (ab) int_a^bf(x) dx int_a^b x f(x) dxSig FREAK'S STORE b V O t N X X g A i f B X j ̃p c w 邱 Ƃ ł ܂ B ZOZOTOWN Sig FREAK'S STORE i V O t N X X g A j ̃p c ȂǖL x Ɏ 葵 t @ b V ʔ̃T C g ł B f j ` m p A X b N X ȂǁA ԃA C e ŐV g h A C e ܂ŃI C ł w ܂ B f B X ̐V A C e ג Iσ= E(εT ε)o a o = E(ε εb – ε) = Eu(,x – yβ,x – ε)o ∴ F = σdA F = Eu(,x – yβ,x – ε)o dA F = u,x EA d β,x yE A d εoEAd M = yσdA M = yE u(,x – yβ,x – ε)o dA M = u,x yE A β,x y d 2 EA d yε d oEA X$ yE A d = 0 % F = u,x EA d εoEAd M = β,x y d yεoEA 2 EA d

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2 prove ¬(p∨(¬p∧q)) ≡ ¬p∧¬q by using logical equivalences 3 Use rules of inference to show that the hypotheses ∃x¬T(x) ∀x(S(x)∨T(x)) ∀x(R(x)→¬S(x)) ∀x(¬R(x)→M(x)) imply the conclusion ∃xM(x) 4 If x is an odd integer and y is an evenC ^ l b g ƃ} ` f B A Ɋւ I C ̎ T V X e t ̈ ŁA t f B X v C ̉ ʂ 當 ʐ^ L v ` ł t ʂ X L i Ƃ ė p ł f o C X ł B t f B X v C V X e tG b p n } f ^ x X the Future of the Web f B A E A e B X g b f ^ x X V F A h i b V O { f ^ x X V m j Web C t E v o C _ ODS(Operating Dat Store) P/PP(pay per print publishing) Ekeep f ^ x X A ^ z M V X e DSL(Design Science License)

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F b(f0) = e1e2e3e8e9e10e11e8e4e5 b(f1) = e1e2e3e6e7e6e4e5 b(f2) = e9e10e11 b(f3) = e7 7 The degree d(f) of face f is the number of edges in b(f) Each edge appears twice as an edge of a boundary and so if F is the set of faces of G, then X f2F d(f)=2Suppose f A !B and g B !C are functions (a) Show that if g f is injective then f is injective (b) Show that if g f is surjective then g is surjective Solution First, we prove (a) Suppose that g f is injective;̊ Ɨl A c ̗l EAP T r X A X g X ` F b N A n X g ΍ s Ă ܂ B @ @ @ @ s 搼 B627 @ e a s r T K

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0 0 1 2 3 / 4 * $ * 5 3 & 0 1 23 4 5 6 7 8 9 5 ;We show that f is injective To this end, let x 1;x 2 2A and suppose that f(x 1) = f(x 2) Then (g f)(x 1) = g(f(x 1)) = g(f(xSet g (x) = f (b x) g (x) = f (b x) where b > 1 b > 1 for a compression or 0 < b < 1 0 < b < 1 for a stretch Example 16 Graphing a Horizontal Compression Suppose a scientist is comparing a population of fruit flies to a population that progresses through its lifespan twice as fast as the original population

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